Integrand size = 85, antiderivative size = 27 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {2 e^4}{7 x \left (1+\frac {3+e^3}{5 \log (x)}\right )} \] Output:
2/7/(1/5*(exp(3)+3)/ln(x)+1)/x*exp(4)
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {10 e^4 \log (x)}{7 x \left (3+e^3+5 \log (x)\right )} \] Input:
Integrate[(E^4*(30 + 10*E^3) + E^4*(-30 - 10*E^3)*Log[x] - 50*E^4*Log[x]^2 )/(63*x^2 + 42*E^3*x^2 + 7*E^6*x^2 + (210*x^2 + 70*E^3*x^2)*Log[x] + 175*x ^2*Log[x]^2),x]
Output:
(10*E^4*Log[x])/(7*x*(3 + E^3 + 5*Log[x]))
Time = 1.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6, 6, 7292, 27, 7292, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-50 e^4 \log ^2(x)+e^4 \left (-30-10 e^3\right ) \log (x)+e^4 \left (30+10 e^3\right )}{7 e^6 x^2+42 e^3 x^2+63 x^2+175 x^2 \log ^2(x)+\left (70 e^3 x^2+210 x^2\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-50 e^4 \log ^2(x)+e^4 \left (-30-10 e^3\right ) \log (x)+e^4 \left (30+10 e^3\right )}{\left (63+42 e^3\right ) x^2+7 e^6 x^2+175 x^2 \log ^2(x)+\left (70 e^3 x^2+210 x^2\right ) \log (x)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-50 e^4 \log ^2(x)+e^4 \left (-30-10 e^3\right ) \log (x)+e^4 \left (30+10 e^3\right )}{\left (63+42 e^3+7 e^6\right ) x^2+175 x^2 \log ^2(x)+\left (70 e^3 x^2+210 x^2\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {10 e^4 \left (-5 \log ^2(x)-3 \left (1+\frac {e^3}{3}\right ) \log (x)+3 \left (1+\frac {e^3}{3}\right )\right )}{7 x^2 \left (5 \log (x)+3 \left (1+\frac {e^3}{3}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {10}{7} e^4 \int \frac {-5 \log ^2(x)-\left (3+e^3\right ) \log (x)+e^3+3}{x^2 \left (5 \log (x)+e^3+3\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {10}{7} e^4 \int \frac {-5 \log ^2(x)-\left (3+e^3\right ) \log (x)+3 \left (1+\frac {e^3}{3}\right )}{x^2 \left (5 \log (x)+3 \left (1+\frac {e^3}{3}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {10}{7} e^4 \int \left (-\frac {1}{5 x^2}+\frac {3+e^3}{5 x^2 \left (5 \log (x)+3 \left (1+\frac {e^3}{3}\right )\right )}+\frac {3+e^3}{x^2 \left (5 \log (x)+3 \left (1+\frac {e^3}{3}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {10}{7} e^4 \left (\frac {1}{5 x}-\frac {3+e^3}{5 x \left (5 \log (x)+e^3+3\right )}\right )\) |
Input:
Int[(E^4*(30 + 10*E^3) + E^4*(-30 - 10*E^3)*Log[x] - 50*E^4*Log[x]^2)/(63* x^2 + 42*E^3*x^2 + 7*E^6*x^2 + (210*x^2 + 70*E^3*x^2)*Log[x] + 175*x^2*Log [x]^2),x]
Output:
(10*E^4*(1/(5*x) - (3 + E^3)/(5*x*(3 + E^3 + 5*Log[x]))))/7
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74
method | result | size |
default | \(\frac {10 \,{\mathrm e}^{4} \ln \left (x \right )}{7 \left (3+5 \ln \left (x \right )+{\mathrm e}^{3}\right ) x}\) | \(20\) |
norman | \(\frac {10 \,{\mathrm e}^{4} \ln \left (x \right )}{7 \left (3+5 \ln \left (x \right )+{\mathrm e}^{3}\right ) x}\) | \(20\) |
parallelrisch | \(\frac {10 \,{\mathrm e}^{4} \ln \left (x \right )}{7 \left (3+5 \ln \left (x \right )+{\mathrm e}^{3}\right ) x}\) | \(20\) |
risch | \(\frac {2 \,{\mathrm e}^{4}}{7 x}-\frac {2 \,{\mathrm e}^{4} \left ({\mathrm e}^{3}+3\right )}{7 x \left (3+5 \ln \left (x \right )+{\mathrm e}^{3}\right )}\) | \(30\) |
Input:
int((-50*exp(4)*ln(x)^2+(-10*exp(3)-30)*exp(4)*ln(x)+(10*exp(3)+30)*exp(4) )/(175*x^2*ln(x)^2+(70*x^2*exp(3)+210*x^2)*ln(x)+7*x^2*exp(3)^2+42*x^2*exp (3)+63*x^2),x,method=_RETURNVERBOSE)
Output:
10/7*exp(4)*ln(x)/(3+5*ln(x)+exp(3))/x
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {10 \, e^{4} \log \left (x\right )}{7 \, {\left (x e^{3} + 5 \, x \log \left (x\right ) + 3 \, x\right )}} \] Input:
integrate((-50*exp(4)*log(x)^2+(-10*exp(3)-30)*exp(4)*log(x)+(10*exp(3)+30 )*exp(4))/(175*x^2*log(x)^2+(70*x^2*exp(3)+210*x^2)*log(x)+7*x^2*exp(3)^2+ 42*x^2*exp(3)+63*x^2),x, algorithm="fricas")
Output:
10/7*e^4*log(x)/(x*e^3 + 5*x*log(x) + 3*x)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {- 2 e^{7} - 6 e^{4}}{35 x \log {\left (x \right )} + 21 x + 7 x e^{3}} + \frac {2 e^{4}}{7 x} \] Input:
integrate((-50*exp(4)*ln(x)**2+(-10*exp(3)-30)*exp(4)*ln(x)+(10*exp(3)+30) *exp(4))/(175*x**2*ln(x)**2+(70*x**2*exp(3)+210*x**2)*ln(x)+7*x**2*exp(3)* *2+42*x**2*exp(3)+63*x**2),x)
Output:
(-2*exp(7) - 6*exp(4))/(35*x*log(x) + 21*x + 7*x*exp(3)) + 2*exp(4)/(7*x)
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {10 \, e^{4} \log \left (x\right )}{7 \, {\left (x {\left (e^{3} + 3\right )} + 5 \, x \log \left (x\right )\right )}} \] Input:
integrate((-50*exp(4)*log(x)^2+(-10*exp(3)-30)*exp(4)*log(x)+(10*exp(3)+30 )*exp(4))/(175*x^2*log(x)^2+(70*x^2*exp(3)+210*x^2)*log(x)+7*x^2*exp(3)^2+ 42*x^2*exp(3)+63*x^2),x, algorithm="maxima")
Output:
10/7*e^4*log(x)/(x*(e^3 + 3) + 5*x*log(x))
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {10 \, e^{4} \log \left (x\right )}{7 \, {\left (x e^{3} + 5 \, x \log \left (x\right ) + 3 \, x\right )}} \] Input:
integrate((-50*exp(4)*log(x)^2+(-10*exp(3)-30)*exp(4)*log(x)+(10*exp(3)+30 )*exp(4))/(175*x^2*log(x)^2+(70*x^2*exp(3)+210*x^2)*log(x)+7*x^2*exp(3)^2+ 42*x^2*exp(3)+63*x^2),x, algorithm="giac")
Output:
10/7*e^4*log(x)/(x*e^3 + 5*x*log(x) + 3*x)
Time = 2.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {10\,{\mathrm {e}}^4\,\ln \left (x\right )}{7\,x\,\left ({\mathrm {e}}^3+5\,\ln \left (x\right )+3\right )} \] Input:
int(-(50*exp(4)*log(x)^2 - exp(4)*(10*exp(3) + 30) + exp(4)*log(x)*(10*exp (3) + 30))/(log(x)*(70*x^2*exp(3) + 210*x^2) + 175*x^2*log(x)^2 + 42*x^2*e xp(3) + 7*x^2*exp(6) + 63*x^2),x)
Output:
(10*exp(4)*log(x))/(7*x*(exp(3) + 5*log(x) + 3))
Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^4 \left (30+10 e^3\right )+e^4 \left (-30-10 e^3\right ) \log (x)-50 e^4 \log ^2(x)}{63 x^2+42 e^3 x^2+7 e^6 x^2+\left (210 x^2+70 e^3 x^2\right ) \log (x)+175 x^2 \log ^2(x)} \, dx=\frac {10 \,\mathrm {log}\left (x \right ) e^{4}}{7 x \left (5 \,\mathrm {log}\left (x \right )+e^{3}+3\right )} \] Input:
int((-50*exp(4)*log(x)^2+(-10*exp(3)-30)*exp(4)*log(x)+(10*exp(3)+30)*exp( 4))/(175*x^2*log(x)^2+(70*x^2*exp(3)+210*x^2)*log(x)+7*x^2*exp(3)^2+42*x^2 *exp(3)+63*x^2),x)
Output:
(10*log(x)*e**4)/(7*x*(5*log(x) + e**3 + 3))